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a) A = \(\frac{101}{19}.\) \(\frac{61}{218}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
= \(\frac{101}{218}.\frac{61}{19}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.\left(\frac{61}{19}-\frac{42}{19}\right)+\frac{117}{218}\)
=\(\frac{101}{218}.\frac{19}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.1+\frac{117}{218}\)
=\(\frac{101}{218}+\frac{117}{218}\)
=\(\frac{218}{218}\)\(=1\)
b) B = \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\)\(.\left(\frac{1}{20}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).0\)
= \(0\)
k chép đề
3/2.A=\(\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+\left(\frac{3}{2}\right)^5+...+\left(\frac{3}{2}\right)^{2013}\)
3/2A-A=(\(\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+\left(\frac{3}{2}\right)^5+...+\left(\frac{3}{2}\right)^{2013}\)) - (\(\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+...+\left(\frac{3}{2}\right)^{2012}\))
1/2 . A =\(\frac{1}{2}+\left(\frac{3}{2}\right)^{2013}\)
A=\(\frac{\frac{1}{2}+\left(\frac{3}{2}\right)^{2013}}{2}\)
B-A=\(\frac{\left(\frac{3}{2}\right)^{2018}}{2}-\)\(\frac{\frac{1}{2}+\left(\frac{3}{2}\right)^{2013}}{2}\)
\(B-A=\frac{\frac{1}{2}}{2}=\frac{1}{2}:2=\frac{1}{4}\)
Câu 3 : \(2+4+6+.........+2n=156\)
\(\Leftrightarrow2\left(1+2+3+.....+n\right)=156\)
\(\Leftrightarrow1+2+3+.........+n=78\)
\(\Leftrightarrow\frac{n\left(n+1\right)}{2}=78\)\(\Leftrightarrow n\left(n+1\right)=156=12.13\)\(\Leftrightarrow n=12\)
Vậy \(n=12\)
a) \(22\frac{1}{2}\cdot\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}\cdot\frac{7}{9}+\frac{50}{100}-\frac{125}{100}\)
\(=\frac{5}{2}\cdot\frac{7}{1}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}=18-\frac{5}{4}=\frac{67}{4}\)
b) \(1,4\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(=\frac{7}{5}\cdot\frac{15}{49}-\frac{22}{15}:\frac{11}{15}\)
\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{22}{15}\cdot\frac{15}{11}\)
\(=\frac{3}{7}-2=\frac{3-14}{7}=\frac{-11}{7}\)
c) \(\left(-\frac{1}{2}\right)^2-\frac{7}{16}:\frac{7}{4}+75\%\)
\(=\frac{1}{4}-\frac{7}{16}\cdot\frac{4}{7}+\frac{75}{100}\)
\(=\frac{1}{4}-\frac{1}{4}+\frac{3}{4}=\frac{3}{4}\)
Bài 2 Bạn tự làm nhé
1.a,\(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{67}{4}\)
b,Các phép tính khác làm tương tự
Đổi các số ra hết thành phân số,có ngoặc thì lm ngoặc trc,Xoq đến nhân chia trước dồi mới cộng trừ
c,tương tự
2.
a,\(1\frac{3}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{8}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{7}{12}\div x=\frac{-77}{20}\)
Đến đây dễ bạn tự làm
b,\(\left(2\frac{4}{5}.x+50\right)\div\frac{2}{3}=-51\)
\(\left(\frac{14}{5}x+50\right)\div\frac{2}{3}=-51\)
\(\frac{14}{5}x+50=-34\)
\(\frac{14}{5}x=-84\)
Tự làm tiếp
c,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)\(\Rightarrow\left|\frac{3}{4}x-\frac{1}{2}\right|=\varnothing\)
\(a)\) \(427-98=329\)
\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)
\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)
\(=30\cdot19+30\cdot43+62\cdot80\)
\(=30\cdot\left(19+43\right)+62\cdot80\)
\(=30\cdot62+62\cdot80\)
\(=62\cdot\left(30+80\right)\)
\(=62\cdot110=6820\)
\(c)\) Đặt \(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2M=1-\frac{1}{3^6}\)
\(\Rightarrow M=\frac{728}{2\cdot729}=\frac{364}{729}\)
Vậy \(M=\frac{364}{729}\)
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2015}\)
\(=\frac{2}{1.2}+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+2+3\right).3}{2}}+.....+\frac{1}{\frac{\left(2015+1\right).2015}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+....+\frac{2}{2015.2016}\)